What is the solution for y^{'''}-2y^{''}-y^'=0 ?

The solution for y^{'''}-2y^{''}-y^'=0 is y=c_{1}+c_{2}e^{(1+sqrt(2))t}+c_{3}e^{(1-sqrt(2))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{'''}-2y^{''}-y^'=0

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} = 0

y = c_{1} + c_{2} e^{(1 + 2) t} + c_{3} e^{(1 - 2) t}

Solution steps

y^{'''} (t) - 2 y^{''} (t) - y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{(1 + 2) t} + c_{3} e^{(1 - 2) t}

y = c_{1} + c_{2} e^{(1 + 2) t} + c_{3} e^{(1 - 2) t}

8 y^{^{''}} + 80 y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

u^{^{''}} + H (t) = 0, H (0) = 0, u^{^{'}} (0) = 2

2 y^{^{''}} + 2 y^{^{'}} + 8 y = 0

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 3, y^{^{'}} (0) = 1

What is the solution for y^{'''}-2y^{''}-y^'=0 ?
The solution for y^{'''}-2y^{''}-y^'=0 is y=c_{1}+c_{2}e^{(1+sqrt(2))t}+c_{3}e^{(1-sqrt(2))t}